Find the limit of (eˣ − 1)/x as x approaches 0.

Step-by-step solution

1. 1

   Check the form. Substituting $x=0$ gives $\dfrac{e^0-1}{0}=\dfrac{0}{0}$, an indeterminate form. Direct substitution is unavailable, so the limit has to be established by one of the standard routes.

2. 2

   Route 1 — recognise a derivative. The difference quotient of $f(x)=e^{x}$ at the point $0$ is exactly this expression:

   $f^{\prime}(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{e^{h}-1}{h}$

   Since $f^{\prime}(x)=e^{x}$ and $f^{\prime}(0)=e^{0}=1$, the limit is $1$. This is the most fundamental view: the limit is the statement that $e^x$ has slope $1$ at the origin.

3. 3

   Route 2 — use the Maclaurin series. From $e^{x}=1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\cdots$,

   $\frac{e^{x}-1}{x}=\frac{x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots}{x}=1+\frac{x}{2}+\frac{x^2}{6}+\cdots\longrightarrow 1$

   This route gives more than the limit: it shows the quotient approaches $1$ linearly, like $1+\tfrac{x}{2}$.

4. 4

   Route 3 — l’Hôpital’s rule. Differentiating top and bottom separately:

   $\lim_{x\to 0}\frac{e^{x}-1}{x}=\lim_{x\to 0}\frac{e^{x}}{1}=1$

   Be aware this is mildly circular if used to prove the result, since computing $\tfrac{d}{dx}e^x$ already relies on this very limit. As a calculation, though, it is immediate and correct.

5. 5

   Confirm numerically and note the one-sided agreement. At $x=0.001$ the quotient is $1.0005002$; at $x=-0.001$ it is $0.9995002$. Both sides approach $1$, and the deviations are $\pm 0.0005\approx\tfrac{x}{2}$, precisely as the series predicted $\checkmark$.